Download 8. Smoothness and the Zariski tangent space We want to give an

8. Smoothness and the Zariski tangent space
We want to give an algebraic notion of the tangent space. In differential geometry, tangent vectors are equivalence classes of maps of
intervals in R into the manifold. This definition lifts to algebraic geometry over C but not over any other field (for example a field of
characteristic p).
Classically tangent vectors are determined by taking derivatives, and
the tangent space to a variety X at x is then the space of tangent
directions, in the whole space, which are tangent to X. Even is this is
how we will compute the tangent space in general, it is still desirable
to have an intrinsic definition, that is a definition which does not use
the fact that X is embedded in Pn .
Now note first that the notion of smoothness is surely local and that
if we want an intrinsic definition, then we want a definition that only
uses the functions on X. Putting this together, smoothness should
be a property of the local ring of X at p. On the other hand taking
derivatives is the same as linear approximation, which means dropping
quadratic and higher terms.
Definition 8.1. Let X be a variety and let p ∈ X be a point of X.
The Zariski tangent space of X at p, denoted Tp X, is equal to the
dual of the quotient
m/m2 ,
where m is the maximal ideal of OX,p .
Note that m/m2 is a vector space. Suppose that we are given a
morphism
f : X −→ Y,
which sends p to q. In this case there is a ring homomorphism
f ∗ : OY,q −→ OX,p
which sends the maximal ideal n into the maximal ideal m. Thus we
get an induced map
df : n/n2 −→ m/m2
On the other hand, geometrically the map on tangent spaces obviously
goes the other way. Therefore it follows that we really do want the
dual of m/m2 . In fact m/m2 is the dual of the Zariski tangent space,
and is referred to as the cotangent space.
In particular, given a morphism f : X −→ Y carrying p to q, then
there is a linear map
df : Tp X −→ Tq Y
1
Definition 8.2. Let X be a quasi-projective variety.
We say that X is smooth at p if the local dimension of X at p is
equal to the dimension of the Zariski tangent space at p.
Now the tangent space to An is canonically a copy of An itself, considered as a vector space based at the point in question. If X ⊂ An ,
then the tangent space to X is included inside the tangent space to An .
The question is then how to describe this subspace.
Lemma 8.3. Let X ⊂ An be an affine variety, of dimension k, and
suppose that f1 , f2 , . . . , fk generates the ideal I of X. Then the tangent
space of X at p, considered as a subspace of the tangent space to An ,
via the inclusion of X in An , is equal to the kernel of the Jacobian
matrix.
Proof. Clearly it is easier to give the dual description of the cotangent
space.
If m is the maximal ideal of OAn ,p and n is the maximal ideal of OX,p ,
then clearly the natural map m −→ n is surjective, so that the induced
map on contangent spaces is surjective. Dually, the induced map on
the Zariski tangent space is injective, so that Tp X is indeed included
in Tp An .
We may as well choose coordinates x1 , x2 , . . . , xn so that p is the
origin. In this case m = hx1 , x2 , . . . , xn i and n = m/I. Moreover m/m2
is the vector space spanned by dx1 , dx2 , . . . , dxn , where dxi denotes
the equivalence class xi + m2 , and n/n2 is canonically isomorphic to
m/(m2 + I). Now the transpose of the Jacobian matrix, defines a linear
map
K k −→ K n = Tp∗ An ,
and it suffices to prove that the image of this map is the kernel of the
map
df : m/m2 −→ n/n2 .
Let g ∈ m. Then
X
g(x) =
ai xi + h(x),
P
where h(x) ∈ m2 . Thus the image of g(x) in m/m2 is equal to i ai dxi .
Moreover, by standard calculus ai is nothing more than
∂g ai =
.
∂xi p
Thus the kernel of the map df is generated by the image of fi in m/m2 ,
which is
X ∂fi dxj ,
∂xj p
j
2
which is nothing more than the image of the Jacobian.
Lemma 8.4. Let X be a quasi-projective variety. Then the function
λ : X −→ N,
is upper semi-continuous, where λ(x) is the dimension of the Zariski
tangent space at x.
Proof. Clearly this result is local on X so that we may assume that X
is affine. In this case the Jacobian matrix defines a morphism π from X
to the space of matrices and the locus where the Zariski tangent space
has a fixed dimension is equal to the locus where this morphism lands
in the space of matrices of fixed rank. Put differently the function λ is
the composition of π and an affine linear function of the rank on the
space of matrices. Since the rank function is upper semicontinuous, the
result follows.
Lemma 8.5. Every irreducible quasi-projective variety is birational to
a hypersurface.
Proof. Let X be a quasi-projective variety of dimension k, with function
field L/K. By assumption there is an intermediary field L/M/K such
that M/K is purely transcendental of transcendence degree k. It follows that the extension L/M is algebraic and finitely generated, whence
it is finite. By the primitive element Theorem, L/M is generated by
one element, say α. It follows that there is polynomial f (x) ∈ M [x]
such that α is a root of f (x). If M = K(x1 , x2 , . . . , xk ), then clearing denominators, we may assume that f (x) ∈ K[x1 , x2 , . . . , xk ][x] '
K[x1 , x2 , . . . , xk+1 ]. But then X is birational to the hypersurface defined by F (X), where F (X) is the homgenisation of f (x).
Proposition 8.6. The set of smooth points of any variety is Zariski
dense.
Proof. Since the dimension of the Zariski tangent space is upper semicontinuous, and always at least the dimension of the variety, it suffices
to prove that every irreducible variety contains at least one smooth
point. By (8.5) we may assume that X is a hypersurface. Passing to
an affine open subset, we may assume that X is an affine hypersurface.
Let f be a definining equation, so that f is an irreducible polynomial.
Then the set of singular points of X is equal to the locus of points
where every partial derivative vanishes. But any partial derivative of
f is a non-zero polynomial of degree one less than f , and so cannot
vanish on X.
3
A basic result in the theory of C ∞ -maps is Sard’s Theorem, which
states that the set of points where a map is singular is a subset of measure zero (of the base). Since any holomorphic map between complex
manifolds is automatically C ∞ , and the derivative of a polynomial is
the same as the derivative as a holomorphic function, it follows that
any morphism between varieties, over C, is smooth over an open subset. In fact by the Lefschetz principle, this result extends to any variety
over C.
Theorem 8.7. Let f : X −→ Y be a morphism of varieties over a field
of characteristic zero.
Then there is a dense open subset U of Y such that if q ∈ U and
p ∈ f −1 (q) ∩ Xsm then the differential dfp : Tp X −→ Tq Y is surjective.
Further, if X is smooth, then the fibres f −1 (q) are smooth if q ∈ U .
Let us recall the Lefschetz principle. First recall the notion of a
first order theory of logic. Basically this means that one describes a
theory of mathematics using a theory based on predicate calculus. For
example, the following is a true statement from the first order theory
of number theory,
∀n∀x∀y∀z n ≥ 3 =⇒ xn + y n 6= z n .
One basic and desirable property of a first order theory of logic is
that it is complete. In other words every possible statement (meaning anything that is well-formed) can be either proved or disproved.
It is a very well-known result that the first order theory of number
theory is not complete (Gödel’s Incompleteness Theorem). What is
perhaps more surprising is that there are interesting theories which are
complete.
Theorem 8.8. The first order logic of algebraically closed fields of
characteristic zero is complete.
Notice that a typical statement of the first order logic of fields is
that a system of polynomial equations does or does not have solution.
Since most statements in algebraic geometry turn on whether or not a
system of polynomial equations have a solution, the following result is
very useful.
Principle 8.9 (Lefschetz Principle). Every statement in the first order
logic of algebraically closed fields of characteristic zero, which is true
over C, is in fact true over any algebraically closed field of characteristic.
4
In fact this principle is immediate from (8.8). Suppose that p is a
statement in the first order logic of algebraically closed fields of characteristic zero. By completeness, we can either prove p or not p. Since
p holds over the complex numbers, there is no way we can prove not
p. Therefore there must be a proof of p. But this proof is valid over
any field of characteristic zero, so p holds over any algebraically closed
field of characteristic zero.
A typical application of the Lefschetz principle is (8.7). By Sard’s
Theorem, we know that (8.7) holds over C. On the other hand, (8.7),
can be reformulated in the first order logic of algebraically closed fields
of characteristic zero. Therefore by the Lefschetz principle, (8.7) is true
over algebraically closed field of characteristic zero.
Perhaps even more interesting, is that (8.7) fails in characteristic p.
Let f : A1 −→ A1 be the morphism t −→ tp . If we fix s, then the
equation
xp − s
is purely inseparable, that is, has only one root. Thus f is a bijection.
However, df is the zero map, since dz p = pz p−1 dz = 0. Thus dfp is
nowhere surjective. Note that the fibres of this map, as schemes, are
isomorphic to zero dimensional schemes of length p.
We now want to aim for a version of the Inverse function Theorem.
In differential geometry, the inverse function theorem states that if a
function is an isomorphism on tangent spaces, then it is locally an
isomorphism. Unfortunately this is too much to expect in algebraic
geometry, since the Zariski topology is too weak for this to be true.
For example consider a curve which double covers another curve. At
any point where there are two points in the fibre, the map on tangent
spaces is an isomorphism. But there is no Zariski neighbourhood of
any point where the map is an isomorphism.
Thus a minimal requirement is that the morphism is a bijection.
Note that this is not enough in general for a morphism between algebraic varieties to be an isomorphism. For example in characteristic
p, Frobenius is nowhere smooth and even in characteristic zero, the
parametrisation of the cuspidal cubic is a bijection but not an isomorphism.
Lemma 8.10. If f : X −→ Y is a projective morphism with finite
fibres, then f is finite.
Proof. Since the result is local on the base, we may assume that Y is
affine. By assumption X ⊂ Y × Pn and we are projecting onto the first
factor. Blowing up a point of Pn , possibly passing to a smaller open
affine subset of Y , we may assume that n = 1, by induction on n.
5
Pick y ∈ Y . Since X/Y has finite fibres, it follows that there is a
point of the fibre {y} × P1 not contained in X. Possibly passing to a
smaller open subset, we may assume that X ⊂ Y × A1 , so that X is
affine. Now X is defined by f (x) ∈ A(Y )[x], where the coefficients of
f (x) lie in A(Y ). Suppose that
f (x) = an xn + an−1 xn−1 + · · · + a0 .
We may always assume that an does not vanish at y. Passing to the
locus where an does not vanish, we may assume that an is a unit, so
that dividing by an , we may assume that an = 1. In this case the ring
B is a quotient of the ring
A[x]/hf i.
But the latter is generated over A by 1, x, . . . xn−1 , and so is a finitely
generated module over A.
Theorem 8.11. Let f : X −→ Y be a projective morphism between
quasi-projective variety varieties.
Then f is an isomorphism iff it is a bijection and the differential dfp
is injective.
Proof. One direction is clear. Otherwise assume that f is projective
and a bijection on closed points. Then f is finite by (8.10). Pick
x ∈ X and let y = f (x). Let A be the local ring of Y at y, B of X at
x. Let φ : A −→ B be the induced ring homomorphism. Then B is a
finitely generated A-module, by assumption, and we just need to show
that φ is an isomorphism.
As f is a bijection on closed points, it follows that φ is injective. So
we might as well suppose that φ is an inclusion. Let m be the maximal
ideal of A and let n be the maximal ideal of B. By assumption
m
n
−→ 2 ,
2
m
n
is surjective. But then
mB + n2 = n.
Nakayama’s Lemma applied to the B-module n/mB it follows that
mB = n. But then
B/A ⊗ A/m = B/(mB + A) = B/(n + A) = 0.
Nakayama’s Lemma applied to the finitely generated A-module B/A
implies that B/A = 0 so that φ is an isomorphism.
Lemma 8.12. Suppose that X ⊂ Pn is a quasi-projective variety and
suppose that π : X −→ Y is the morphism induced by projection from
a linear subspace.
6
Let y ∈ Y . Then π −1 (y) = hΛ, yi ∩ X. If further this fibre consists of
one point, then the map between Zariski tangent spaces is an isomorphism if the intersection of hΛ, xi with the Zariksi tangent space to X
at x has dimension zero.
Proof. Easy.
Lemma 8.13. Let X be a smooth irreducible subset of Pn of dimension
k. Consider the projection Y of X down to a smaller dimensional
projective space Pm , from a linear space Λ of dimension n − m − 1.
If the dimension of m ≥ 2k + 1 and Λ is general (that is belongs to
an appropriate open subset of the Grassmannian) then π is an isomorphism.
Proof. Since projection from a general linear space is the same as a
sequence of projections from general points, we may assume that Λ is
in fact a point p, so that m = n − 1.
Now we know that π is a bijection provided that p does not lie on any
secant line. Since the secant variety has dimension at most 2k + 1, it
follows that we may certainly find a point away from the secant variety,
provided that n > 2k + 1. Now since a tangent line is a limit of secant
lines, it follows that such a point will also not lie on any tangent lines.
But then π is then an isomorphism on tangent spaces, whence an
isomorphism.
For example, it follows that any curve may be embedded in P3 and
any surface in P5 . Now let us turn to the following classical problem
in enumerative geometry.
Question 8.14. Let C ⊂ Pn be a curve in P2 and let p ∈ P2 .
How many tangent lines does p lie on?
The first thing that we will need is a natty way to describe the
projective tangent space to a variety.
Definition 8.15. Let X ⊂ Pn .
The projective tangent space to X at p is the closure of the
affine tangent space.
In other words the projective tangent space has the same dimension as the affine tangent space and is obtained by adding the suitable
points at infinity. Suppose that the curve is defined by the polynomial
F (X, Y, Z). Then the tangent line to C at p, is
∂F ∂F ∂F X+
Y +
Z.
∂X ∂Y ∂Z p
p
7
p
Of course it suffices to check that we get the right answer on an affine
piece.
Lemma 8.16. Let F be a homogeneous polynomial of degree d in
X0 , X1 , . . . , Xn . Then
X
∂F
dF =
Xi
∂Xi
Proof. Both sides are linear in F . Thus it suffices to prove this for a
monomial of degree d, when the result is clear.
It follows then that the tangent line above does indeed pass through
p. The rest is easy.
Finally we will need Bézout’s Theorem.
Theorem 8.17 (Bézout’s Theorem). Let C and D be two curves defined by homogenous polynomials of degrees d and e. Suppose that C∩D
does not contain a curve.
Then |C ∩ D| is at most de, with equality iff the intersection of the
two tangent spaces at p ∈ C ∩ D is equal to p.
We are now ready to answer (8.14).
Lemma 8.18. Let C ⊂ Pn be a curve in P2 and let p ∈ P2 be a general
point.
Then p lies on d(d − 1) tangent lines.
Proof. Fix p = [a : b : c] and let D be the curve defined by
∂F
∂F
∂F
G=a
+b
+c
.
∂X
∂Y
∂Z
Then G is a polynomial of degree d − 1. Consider a point q where C
intersects D. Then the tangent line to C at q is given by
∂F ∂F ∂F X+
Y +
Z.
∂X ∂Y ∂Z q
q
q
But then since p satisfies this equation, as q lies on D, it follows that
p lies on the tangent line of C at q. Similarly it is easy to check the
converse, that if p lies on the tangent line to C at q, then q is an
intersection point of C and D.
Now apply Bézout’s Theorem.
There is an interesting way to look at all of this. In fact one may
generalise the result above to the case of curves with nodes. Note that
if you take a curve in P3 and take a general projection down to P2 , then
you get a nodal curve. Indeed it is easy to pick the point of projection
not on a tangent line, since the space of tangent lines obviously sweeps
8
out a surface; it is a little more involved to show that the space of
three secant lines is a proper subvariety. (8.18) was then generalised to
this case and it was shown that if δ is the number of nodes, then the
number
d(d − 1)
−δ
2
is an invariant of the curve.
Here is another way to look at this. Suppose that we project our
curve down to P1 from a point. Then we get a finite cover of P1 , with d
points in the general fibre. Lines tangent to C passing through p then
count the number of branch points, that is, the number of points in the
base where the fibre has fewer than d points. Since this tangent line
is only tangent to p and is simply tangent (that is, there are no flex
points) there are d − 1 points in this fibre, and the ramification point
corresponding to the branch point is where two sheets come together.
The modern approach to this invariant is quite different. If we are
over the complex numbers C, changing perspective, we may view the
curve C as a Riemann surface covering another Riemann surface D.
Now the basic topological invariant of a compact oriented Riemann
surface is it’s genus. In these terms there is a simple formula that
connects the genus of C and B, in terms of the ramification data,
known as Riemann-Hurwitz,
2g − 2 = d(2h − 2) + b,
where g is the genus of C, h the genus of B, d the order of the cover
and b the contribution from the ramification points. Indeed if locally
on C, the map is given as z −→ z e so that e sheets come together, the
contribution is e − 1.
In our case, B = P1 which is of genus 0, for each branch point, we
have simple ramification, so that e = 2 and the contribution is one,
making a total b = d(d − 1). Thus
2g − 2 = −2d + d(d − 1).
Solving for g we get
(d − 1)(d − 2)
.
2
Note that if d ≤ 2, then we get g = 0 as expected (that is C ' P1 ) and
if d = 3 then we get an elliptic curve.
It also seems worth pointing out that if we take a smooth variety X
and blow up a point p, then the exceptional divisor E is canonically
the projectivisation of the Zariski tangent space to X at p,
g=
E = P(Tp X).
9
Indeed the point is that E picks up the different tangent directions to
X at p, and this is exactly the set of lines in Tp X. Note the difference
between the projective tangent space and the projectivisation of the
tangent space.
It also seems worth pointing out that one define the Zariski tangent
space to a scheme X, at a point x, using exactly the same definition,
the dual of
m/m2 ,
where m ⊂ OX,x is the maximal ideal of the local ring. We have already
seen that this the same as looking at the space of maps
Hom(Spec k[]/h2 i, X),
which sends hi to x and where k is the residue field of x ∈ X. However
in general, if we have the equality of dimensions of both the Zariski
tangent space and the local dimension, we only call X regular at
x ∈ X. Smoothness is a more restricted notion in general.
Having said this, if X is a quasi-projective variety over an algebraically closed field then X is smooth as a variety if and only if it is
smooth as a scheme over Spec k. In fact an abstract variety over Spec k
is smooth if and only if it is regular. Note that if x is a specialisation
of ξ and X is regular at x then X is regular at ξ, so it is enough to
check that X is regular at the closed points.
Note that one can sometimes use the Zariski tangent space to identify
embedded points. If X is a scheme and Y = Xred is the reduced
subscheme then x ∈ X is an embedded point if
dim Tx X > dim Tx Y,
and X is reduced out of x. For example, if X is not regular at x but Y
is regular at x then x ∈ X is an embedded point. Note however that
it is possible that x ∈ X is an embedded point but the Zariski tangent
space is no bigger than it should be; for example if H03 is the punctual
Hilbert scheme of a smooth surface then the underlying variety is a
quadric cone. The vertex of the cone corresponds to the unique zero
dimensional length three scheme which is not curivilinear and this is
an embedded point, even though the Zariski tangent space is three
dimensional.
It is interesting to see which toric varieties are smooth. The question
is local, so we might as well assume that X = Uσ is affine. If σ ⊂ NR
does not span NR , then X ' Uσ0 × Glm , where σ 0 is the same cone as
σ embedded in the space it spans. So we might as well assume that
σ spans NR . In this case X contains a unique fixed point xσ which
is in the closure of every orbit. Since X only contains finitely many
10
orbits, it follows that X is smooth if and only if X is regular at xσ .
The maximal ideal of xσ is generated by χu , where u ∈ Sσ . The square
of the maximal is generated by χu+v , where u and v are two elements
of Sσ . So a basis for m/m2 is given by elements of Sσ that are not sums
of two elements. Since the elements of Sσ generate the group M , the
elements of Sσ which are not sums of two elements, must generate the
group. Given an extremal ray of σ̌, a primitive generator of this ray is
not the sum of two elements in Sσ . So σ̌ must have n edges and they
must generate M . So these elements are a basis of the lattice and in
fact X ' Ank .
11